大师作文网化简求值 4(x平方-x)除以x-1加(x-2)的平方,其中x=根号七
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/12 08:17:32
因为x=1+根号2y=1减根号2所以xy=1+根号2乘以1减根号2=1-2=-1x+y=2所以可以求得,原式=xy/(x+y)=-1/2
原式=(x³-4x)-(x³-3x²+2x-3x²-9x+6)-(2x²-4x)=x³-4x-x³+7x-6-2x²+4
是(x^2/(x-1)-x+1)/(4x^2/(1-x)-4x+1)吗原式=(x-x+1)(x+x-1)/(x-1)/((4x^2-4x+1+4x^2-x)/(1-x))=-(2x-1)/(4x^2-
叙述好混乱,看不明白,请写成{[(a+b)/(c-d)]/(e+f)}+[(g-h)/i]之类的表达式
(xy-x的平方）除以xy分之x的平方-2xy+y的平方乘x的平方分之x-y=x(y-x)*xy/(x-y)²*（x-y）/x²=-y再问：不好意思，打错了(xy-y的平方）除以x
(xy-x²)÷(x²-2xy+y²)/(x-y)×(x-y)/x²=-x(x-y)÷(x-y)²/x²=-x(x-y)×x²/(
x平方-2x/x平方-1除以[(x-1)-(2x-1/x+1)]=[x(x-2)/(x+1)(x-1)]/[(x^2-1-2x+1)/(x+1)]=x(x-2)/(x+1)(x-1)×(x+1)/x(
(4x²-3x)+(2+4x-x²)-(2x²+x+1)先去括号；4x²-3x+2+4x-x²-2x²-x-1合并同类项x²+1代
{(x^2-6x+9)/(x^2+4x+4)}/{(x^2-9)/x}=(x-3)^2/(x+2)^2*x/{(x+3)(x-3)}=(x-3)/(x+2)^2*x/(x+3)=x(x-3)/{(x+
原式=[(x-1+1)/(x-1)]*[(x^2-1)/x]=[x/(x-1)]*[(x+1)(x-1)/x]=x+1=-2+1=-1
/>原式=[y/(x-y)]+[y³/(x³-2x²y+xy²)]÷[(xy+y²)/(y²-x²)]=[y/(x-y)]+{y&
分子分母同乘以4.再一化简
再答：满意吗，亲，别忘了给好评哦
:(x^2-4x)/(x+1)除以(4-x)/x,=x(x-4)/(x+1)*x/(4-x)=-x^2/(x+1)=-9/(-2)=9/2
=(x²+x-x)/(x+1)×(x+1)/(x+2)(x+1)×(x+2)(x-2)/x(x-2)=x²/(x+1)×(x+1)/(x+2)(x+1)×(x+2)(x-2)/x(